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parabola y=ax^2+k has vertex (0,-1) and passes through the point (3,3). find the equation

 Mar 3, 2019

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 #1
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\(\text{if we've got vertex at }(0,-1) \text{ then}\\ y = a x^2-1\\ \text{we're told that }(3,3) \text{ is on the parabola}\\ 3 = a 3^2 - 1\\ 4 = 9a\\ a = \dfrac{4}{9}\\ y = \dfrac 4 9 x^2 - 1\)

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 Mar 3, 2019
 #1
avatar+6248 
+1
Best Answer

\(\text{if we've got vertex at }(0,-1) \text{ then}\\ y = a x^2-1\\ \text{we're told that }(3,3) \text{ is on the parabola}\\ 3 = a 3^2 - 1\\ 4 = 9a\\ a = \dfrac{4}{9}\\ y = \dfrac 4 9 x^2 - 1\)

Rom Mar 3, 2019
 #2
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thank you for your help

Guest Mar 3, 2019

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